Chapter 8: Rotation Equilibrium and Rotational Dynamics

transcript

PHU 205 Mechanics for Life Sciences

Chapter 8:

Rotation Equilibrium and Rotational Dynamics

Prof. Liliana Braescu & Prof. Nouredine Zettili

Required text College Physics Raymond A. Serway and Chris Vuille 9th Edition, 2012 BROOKS/COLE CENGAGE Learning ISBN 10: 1-111-42745-3 ISBN 13: 978-1-111-42745-0

Course Outline Chapter 8 Rotational Equilibrium and Rotational

Dynamics 8.1 Torque 8.2 Torque and the Two Conditions for Equilibrium 8.3 The Center of Gravity 8.4 Examples of Objects in Equilibrium 8.5 Relationship Between Torque and Angular

Acceleration 8.6 Rotational Kinetic Energy 8.7 Angular Momentum 2

Chapter 8 Rotational Equilibrium and Rotational

Acceleration 8.6 Rotational Kinetic Energy 8.7 Angular Momentum

8.1 Torque

Force versus Torque u Forces cause accelerations u Torques cause angular accelerations u Force and torque are related u torque – the tendency of a force to rotate an object about a given axis (see figure: top view of a door)

τ = Fd

Torque of a force perpendicular to the lever arm d

8.1 Torque u Magnitude of the torque (vector) – exerted by the

force F which makes an angle with the lever arm r:

τ = F ⋅ r sinθ

F = F , r = r are magnitudes of vectors F and r

8.1 Torque

Remarks: u If => no rotation u If => maximum torque u If => no lever arm

u Sign convention for torque: ✔ force causes a clockwise rotation ✔ force causes a counterclockwise rotation u SI unit of torque → Nm

0 0θ τ= ⇒ =90

2Frπθ τ= ° = ⇒ =

0 0r τ= ⇒ =

0τ <0τ >

8.1 Torque

General Definition of Torque u When the force is parallel to the position vector, no rotation occurs: torque=0 u When the force is at some angle, the perpendicular component causes the rotation

8.1 Torque

Multiple Torques u When two or more torques are acting on an object, the torques are added u If the net torque is zero, the object’s rate of rotation doesn’t change.

8.2 Torque and the Two Conditions for Equilibrium

u  Rotational equilibrium – an object is in rotational

equilibrium if the net torque about any axis is zero: u  Static equilibrium – an object is in static

equilibrium if the following two conditions are jointly satisfied:

✔ The net external force is zero: ✔ The net torque is zero:

0τ =∑

0F =∑0τ =∑

8.3 The Center of Gravity u Consider a number of masses

(m1, m2, m3,…) located at (x1,y1), (x2,y2),(x3,y3)…

u The coordinates of the center of gravity of this object are as follows:

1 1 2 2 3 3

xx x xxmm m m

m m m m+ += =+ +

∑∑

1 1 2 2 3 3

m ym y m y m yym m m m+ += =+ +

∑∑

Problems u Problem 8.1

Four objects are situated along the y axis as follows: a 2.00-kg object is at +3.00 meters, a 3.00-kg object is at +2.50 meters, a 2.50-kg object is at the origin, and a 4.00-kg object is at -0.500 meters. Question: Where is the center of gravity of this system?

Problems u Problem 8.1 Solution

Ycg = Σmiyi/Σmi =[ 2.00(3.00) + 3.00(2.50) + 2.50(0) +4.00(-0.50)] / [2.00 + 3.00+ 2.50 + 4.00] = 1.00 m. Xcg = Σmixi/Σmi=0 m because objects are situated along y axis.

8.4 Examples of Objects in Equilibrium u Rotational equilibrium: u Static equilibrium:

0τ =∑

0F =∑0τ =∑ Beam:

✔The free body diagram includes the directions of the forces ✔The weights act through the centers of gravity of their objects

8.4 Examples of Objects in Equilibrium

Ladder: ✔The free body diagram shows the normal force and the force of static friction acting on the ladder at the ground ✔The last diagram shows the lever arms for the forces

A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 200 N and is 3.00 meters long. Question: What is the tension in each rope when the 700-N worker stands 1.00 meter from one end? Hint: Free body diagram

200 N1.5 m

1 m 2 mT2

Problems u Problem 8.2 Solution Static equilibrium: ,

u Taking torques about the left end of the scaffold, we have: T1(0) - (700 N)(1 m)

– (200 N)(1.50 m) + T2(3.00 m) = 0. From this, T2 = 333 N. u Then, from ΣFy = 0, we have

T1 + T2 - 700 N - 200 N = 0. u Since T2 = 333 N, we find T1 = 567 N.

200 N1.5 m

1 m 2 mT2

0F =∑0τ =∑ τ = Fd

A 20.0-kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole, as shown in the figure. A cable at an angle of 30.0° with the beam helps to support the light. Questions: (a) the tension in the cable and (b) the horizontal and vertical forces exerted on the beam by the pole.

200 Nd

PivotHint: Free body diagram

(a) From Στ = 0 about the pivot shown, we have (T sin30°)d - (200 N)d = 0, where d is the length of the beam. The forces H, V, and Tx produce no torque about the pivot point. Solving for T, we find T = 400 N.

200 Nd

0F =∑0τ =∑ τ = Fd

(b) From ΣFx = 0, we have

H – T cos30.0° = 0. Because T = 400 N => H = 346 N (to right). From ΣFy = 0, we have V + Tsin30.0° - 200 N = 0, which yields V = 0.

200 Nd

0F =∑0τ =∑ τ = Fd

8.5 Relationship Between Torque and Angular Acceleration

u Consider a mass m which is rotated on a frictionless horizontal surface with a force F. u According to Newton’s second law, u Multiplying both sides by r gives

tF ma=

tFr mra=2

2cva rr

ω= = at= αr

u This equation is Newton’s second law applied to rotational motion: u From this, we define the moment of inertia of a mass m with respect to the point o as u SI unit of the moment of inertia → kgm2

at = rα ⇒ Fr

! = mr2αIoα! o oIτ α=

2oI mr=

More About Moment of Inertia u The moment of inertia I is the counterpart of mass in rotational motion. u The moment of inertia depends on: (a)  the quantity of matter, (b)  its distribution in the rigid object, and (c) upon the location of the axis of rotation

Examples of Moment of Inertia

Moment of inertia of three masses m1, m2 and m3 rotating about an axis and located at r1, r2 and r3 from the axis is given by

2 2 21 2 3 1 1 2 2 3 3o II I I m r m r m r+ += + = +

Examples of Moment of Inertia

u Image the hoop is divided into a number of small segments, m1 … u These segments are equidistant from the axis where I = Σmiri

2 = MR2

1 2 3 4 ...M m m m m= + + + +

Other Examples of Moment of Inertia

I = Σmiri

I = ρ r( )r2 dV

This rod (dimension 1D) ρ - Mass Density m – mass, m=ρsl or ρsL

Remarks: u What is the physical meaning of the moment of inertia?

✔Consider two hoops of the same mass M and radius R — one made of steel and the other made of wood. ✔I depends only on the mass and size (geometry) of the object; therefore, they have the same moment of inertia. ✔An object with a small moment of inertia is easier to set into rotation than another with a larger moment of inertia.

Four masses are held in position at the corners of a rectangle by light rods, as shown in the figure. Questions: Find the moment of inertia of the system about (a) the x axis, (b) the y axis, and (c) an axis through O and perpendicular to the page.

(a) First, apply the above equation about the x axis. We have, Ix = (3.00 kg)(9.00 m2) + (2.00 kg)(9.00 m2) +(2.00 kg)(9.00 m2) + (4.00 kg)(9.00 m2) = = 99.0 kg m2.

I = Σmiri2

(b) About the y axis we have: Iy = (3.00 kg)(4.00 m2) + (2.00 kg)(4.00 m2) + (2.00 kg)(4.00 m2) + (4.00 kg)(4.00 m2) = = 44.0 kg m2.

I = Σmiri2

(c) About O perpendicular on the page Io = (3.00 kg)(13 m2) + (2.00 kg)(13 m2) + +(2.00 kg)(13 m2) + (4.00 kg)(13 m2) = (11.00 kg)(13.00 m2) = 143 kg m2.

I = Σmiri2

r1 = r2 = r3 = r4 = 32 + 22 = 13

8.6 Rotational Kinetic Energy u An object with a moment of inertia I rotating at an angular speed ω about a fixed axis has a rotational kinetic energy of u An object with a moment of inertia I simultaneously rotating at angular speed ω and rolling at velocity v will have a kinetic energy consisting of two parts - translational and rotational

KErot =

KEtot = KEtrans + KErot

KEtot =12

mv2 + 12

8.6 Rotational Kinetic Energy u If we have conservative forces only (no dissipative forces such as friction can be present), total energy is conserved (mechanical energy is const. ) u If dissipative forces are present then instead of conservation of energy we have Work-Energy Theorem (see §5.5):

PEi + KEi = PE f + KE f

PEi + KEroti+ KEtransi

= PE f + KErot f+ KEtrans f

Wnet =WC

Wnet =WC +WNC

WNC = ΔEnergy = ΔKEtrans + ΔKErot + ΔPE

A 10.0-kg cylinder rolls without slipping on a rough surface. At the instant its center of mass has a speed of 10.0 m/s. Questions: Determine (a) the translational kinetic energy of its center of mass, (b) the rotational kinetic energy about its center of mass, and (c) its total kinetic energy.

(a) KEtrans = (1/2)mv2 = (1/2)(10.0 kg)(10.0 m/s)2 = 500 J.

(b) KErot = (1/2)Iω2

= (1/2)(1/2)mr2(v2/r2) = (1/4)(10.0 kg)(10.0 m/s)2 = 250 J.

(c) KEtotal = KEtrans + KErot = 750 J.

8.7 Angular Momentum u An object with a moment of inertia I rotating at an angular speed ω has an angular momentum given by u SI unit of angular momentum → Js u If i.e., angular momentum is conserved.

L Iω=

τ = 0⇒ ΔL = 0⇒ Li = Lf ⇒ Iiω i = I fω f

A cylinder with a moment of inertia I1 rotates with angular velocity ωo about a frictionless vertical axis. A second cylinder, with moment of inertia I2, initially not rotating, drops onto the first cylinder (figure below). Because the surfaces are rough, the two eventually reach the same angular velocity, ω.

Questions: (a) Calculate ω. (b) Show that kinetic energy is lost in this situation, and calculate the ratio of the final to the initial kinetic energy.

(a) From conservation of angular momentum:

(I1 + I2)ω = I1ωo, or ω=(I1/(I1+I2))ωo.

(b) Kf = (1/2)(I1+I2)ω2, and Ki = (1/2)I1ωo2

So, after some algebra, Kf /Ki = I1/(I1+I2),

which is less than 1.

Chapter 8 Ends

Quiz 7-8 is coming ….. before Chapter 9 For reviewing Chapter 8, Study the following problems: 8.1 page 237; 8.2 page 239; 8.4

; 8.7 page 245; 8.8 page 246; 8.9 page 249; 8.12 page 255; 8.13 page 255.

Solve the following problems: 3, 8 page 265; 17 page 266; 18, 21, 22 page 267; 30 page 268, 46 page 270; 63 page 272.

Textbook “College Physics” Raymond A. Serway and Chris Vuille 9th Edition, 2012

Chapter 8: Rotation Equilibrium and Rotational Dynamics

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